Question

A transverse wave is moving along a string. The linear density of a vibrating string is \(2 \times 10^{-3} \mathrm{~kg} / \mathrm{m}\). The wave equation is \(y=0.02 \sin (2 x+10 t) m\). The tension in the string is

• A. \(2\times 10^{-2}\text{ }\mathrm{N}\)
• B. \(5\times 10^{-2}\text{ }\mathrm{N}\)
• C. \(2\times 10^{-3}\text{ }\mathrm{N}\)
• D. \(5\times 10^{-3}\text{ }\mathrm{N}\)

Hint:

Read the wave speed from the equation as omega over k, then use tension equals linear density times speed squared.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept
Compare \(y = 0.02\sin(2x + 10t)\) with \(y = A\sin(kx + \omega t)\): \(k = 2\ \text{m}^{-1}\), \(\omega = 10\ \text{rad/s}\).

Step 2: Wave Speed
\(v = \dfrac{\omega}{k} = \dfrac{10}{2} = 5\ \text{m/s}\)

Step 3: Tension
\(v = \sqrt{T/\mu}\Rightarrow T = \mu v^2 = 2\times10^{-3}\times 25\)

Step 4: Result
\(T = 5\times10^{-2}\ \text{N}\), option B.

Final Answer:
The tension in the string is 5 times 10 to the power minus 2 newton. \[ \boxed{\text{(B) }5\times 10^{-2}\ \text{N}} \]